Since both partial derivatives $\pdiff{f}{x}(x,y)$ and $\pdiff{f}{y}(x,y)$ are continuous functions,
we know that $f(x,y)$ is differentiable.
Therefore, $Df(1,2)$ is the derivative of $f$, and the function has a tangent plane there.

To calculate the equation of the tangent plane, the only additional calculation is the value of $f$ at $(x,y)=(1,2)$, which is $f(1,2) = 1^2+2^2=5$. The equation for the tangent plane is
\begin{align*}
z &= f(1,2)+\pdiff{f}{x}(1,2)(x-1) + \pdiff{f}{y}(1,2)(y-2) \\
&= 5 + 2(x-1) + 4(y-2)
\end{align*}

For a scalar-valued function of two variables such as $f(x,y)$, the
tangent plane is the linear approximation. We can write the
linear approximation as
\begin{align*}
L(x,y) = 5 + 2(x-1) + 4(y-2).
\end{align*}

Note that $(1.1,1.9,0.1)$ is very close to $(1,2,0)$, which is the
point around which we computed the linear approximation. So, we
expect this linear approximation to be close to the true value of
$\vc{f}$ at $(1.1,1.9,0.1)$. Let's compare the above answer to the
actual value of $\vc{f}$ at $(1.1,1.9,0.1)$:
\begin{align*}
\vc{f}(1.1,1.9,0.1)
&= ((1.1)^2(1.9)^2(0.1), 1.9+\sin (0.1))\\
&\approx (0.4368,1.9998).
\end{align*}
In this case, the approximation is quite close.